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Representation of Numbers 263
For the negative numbers we have the same formats as the fixed-point
representations, including the 10s-complement format.
The number of digits used in the exponent determine the range of
representable numbers, whereas the number of digits used in the mantissa
determine the precision of the numbers. For example, if the mantissa is
expressed using 2 digits plus the sign, and the exponent is expressed using
2 digits plus the sign, then the real number line will be covered as:
99 -99 -99 99
-9.99x10 -1.0x10 0 1.0x10 9.99x10
----------+-------------+-----------+-----------+-------------+----------->
| accessible | 0 | accessible |
negative | negative | negative | positive | positive | positive
overflow | numbers | underflow | underflow | numbers | overflow
The range of accessible floating-point numbers with a given representation
can be large, but it is still finite. In the preceding example (e.g., with 2 digits
for the mantissa and 2 digits for the exponent), there are only 9 × 10 ×
10×199 = 179, 100 positive numbers, and as many negative numbers, plus
the number zero, for a total of 358,201 numbers that can be represented.
Binary numbers Although the fraction-only fixed-point arithmetic
does not have any overflow problems when two numbers are multiplied, it
does suffer from overflow problems when two numbers are added. Also, the
fixed-point numbers have limited dynamic range. Both of these aspects
are unacceptable for an intensive computational job. These limitations
can be removed by making the binary point floating rather than fixed.
The floating-point bit arrangement for binary-number representation
is similar to that for the decimal numbers. In practice, however, two exceptions
are made. The exponent is expressed using L-bit excess-2 L−1
format, and the B-bit normalized mantissa is a fractional number with
a1following the binary point. Note that the sign bit is the MSB of the
bit pattern. Thus the B-bit mantissa and L-bit exponent (for a total of
B + L +1word length) bit pattern is given by (note the reversal of the
mantissa exponent places)
ˆx =
Sign of M
↓
± xx ···x
} {{ }
L-bit E
1x ···x
} {{ }
B-bit M
(6.42)
where exponent E is adjusted so that we have a normalized mantissa—
that is, 1/2 ≤ M